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∵a1=25,S9=S17,∴9a1+9×82d=17a1+17×162d,解得d=-2.∴Sn=25n+×n(n−1)2(-2)=-n2+26n=-(n-13)2+169.由二次函數的知識可知:當n=13時,S13=169,即前13項之和最大,最大值為169.故答案為:13
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